What is the integral of 4/(x-3) ?

The integral of 4/(x-3) is 4ln|x-3|+C

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integral of 4/(x-3)

\int \frac{4}{x - 3} d x

4 ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{4}{x - 3} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int \frac{1}{x - 3} d x

Apply u-substitution

: \int \frac{1}{u} d u

= 4 \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 ln ∣ u ∣

Substitute back

u = x - 3

= 4 ln ∣ x - 3 ∣

Add a constant to the solution

= 4 ln ∣ x - 3 ∣ + C

\frac{d}{d x} (x sin (x) + x cos (x))

n = 0 \sum \infty (\frac{7}{4})^{n}

x \to \infty + lim (\frac{1}{5 x})

f (x) = 3 x + 40, at x = 3

\int (\frac{3 e ^{- x} + 2 e ^{3 x}}{e ^{2 x}}) d x